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the work week for adults in the us that work full time is normally dist…

Question

the work week for adults in the us that work full time is normally distributed with a mean of 47 hours. a newly hired engineer at a start - up company believes that employees at start - up companies work more on average then most working adults in the us. she asks 12 engineering friends at start - ups for the lengths in hours of their work week. their responses are shown in the table below. test the claim using a 1% level of significance. give answer to at least 4 decimal places.
hours
48
47
59
50
49
63
47
45
47
46
52
51

Explanation:

Step1: Calculate sample mean $\bar{x}$

$\bar{x}=\frac{48 + 47+59+50+49+63+47+45+47+46+52+51}{12}=\frac{594}{12}=49.5$

Step2: Calculate sample standard - deviation $s$

First, calculate the squared differences from the mean for each data - point:
$(48 - 49.5)^2=(-1.5)^2 = 2.25$, $(47 - 49.5)^2=(-2.5)^2 = 6.25$, $(59 - 49.5)^2=(9.5)^2 = 90.25$, $(50 - 49.5)^2=(0.5)^2 = 0.25$, $(49 - 49.5)^2=(-0.5)^2 = 0.25$, $(63 - 49.5)^2=(13.5)^2 = 182.25$, $(47 - 49.5)^2=(-2.5)^2 = 6.25$, $(45 - 49.5)^2=(-4.5)^2 = 20.25$, $(47 - 49.5)^2=(-2.5)^2 = 6.25$, $(46 - 49.5)^2=(-3.5)^2 = 12.25$, $(52 - 49.5)^2=(2.5)^2 = 6.25$, $(51 - 49.5)^2=(1.5)^2 = 2.25$
The sum of squared differences $\sum_{i = 1}^{n}(x_i-\bar{x})^2=2.25 + 6.25+90.25+0.25+0.25+182.25+6.25+20.25+6.25+12.25+6.25+2.25 = 335$
$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{335}{11}}\approx5.5055$

Step3: State the hypotheses

$H_0:\mu\leq47$ (null hypothesis), $H_1:\mu>47$ (alternative hypothesis)

Step4: Calculate the test - statistic

The test - statistic for a one - sample t - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\mu = 47$, $\bar{x}=49.5$, $s\approx5.5055$, and $n = 12$
$t=\frac{49.5 - 47}{5.5055/\sqrt{12}}\approx1.5675$

Step5: Determine the critical value

The degrees of freedom is $df=n - 1=12 - 1 = 11$. For a one - tailed test with $\alpha = 0.01$, the critical value $t_{\alpha,df}=t_{0.01,11}= 2.7181$

Step6: Make a decision

Since $t = 1.5675

Answer:

We fail to reject the null hypothesis. The test - statistic is approximately $t = 1.5675$ and the critical value is $t_{0.01,11}=2.7181$.